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CHAPTER 4 4.1. The value of E at P (ρ = 2, φ = 40◦ , z = 3) is given as E = 100aρ − 200aφ + 300az V/m. Determine the incremental work required to move a 20 µC charge a distance of 6 µm: a) in the direction of aρ : The incremental work is given by dW = −q E · dL, where in this case, dL = dρ aρ = 6 × 10−6 aρ . Thus dW = −(20 × 10−6 C)(100 V/m)(6 × 10−6 m) = −12 × 10−9 J = −12 nJ b) in the direction of aφ : In this case dL = 2 dφ aφ = 6 × 10−6 aφ , and so dW = −(20 × 10−6 )(−200)(6 × 10−6 ) = 2.4 × 10−8 J = 24 nJ c) in the direction of az : Here, dL = dz az = 6 × 10−6 az , and so dW = −(20 × 10−6 )(300)(6 × 10−6 ) = −3.6 × 10−8 J = −36 nJ d) in the direction of E: Here, dL = 6 × 10−6 aE , where aE =

100aρ − 200aφ + 300az = 0.267 aρ − 0.535 aφ + 0.802 az [1002 + 2002 + 3002 ]1/2

Thus dW = −(20 × 10−6 )[100aρ − 200aφ + 300az ] · [0.267 aρ − 0.535 aφ + 0.802 az ](6 × 10−6 ) = −44.9 nJ e) In the direction of G = 2 ax − 3 ay + 4 az : In this case, dL = 6 × 10−6 aG , where aG =

2ax − 3ay + 4az = 0.371 ax − 0.557 ay + 0.743 az [22 + 32 + 42 ]1/2

So now dW = −(20 × 10−6 )[100aρ − 200aφ + 300az ] · [0.371 ax − 0.557 ay + 0.743 az ](6 × 10−6 )
= −(20 × 10−6 ) 37.1(aρ · ax ) − 55.7(aρ · ay ) − 74.2(aφ · ax ) + 111.4(aφ · ay ) + 222.9] (6 × 10−6 ) where, at P , (aρ · ax ) = (aφ · ay ) = cos(40◦ ) = 0.766, (aρ · ay ) = sin(40◦ ) = 0.643, and (aφ · ax ) = − sin(40◦ ) = −0.643. Substituting these results in dW = −(20 × 10−6 )[28.4 − 35.8 + 47.7 + 85.3 + 222.9](6 × 10−6 ) = −41.8 nJ

42

4.2. Let E = 400ax − 300ay + 500az in the neighborhood of point P (6, 2, −3). Find the incremental work done in moving a 4-C charge a distance of 1 mm in the direction specified by: a) ax + ay + az : We write dW = −qE · dL = −4(400ax − 300ay + 500az ) · =−

(ax + ay + az ) (10−3 ) √ 3

(4 × 10−3 ) (400 − 300 + 500) = −1.39 J √ 3

b) −2ax + 3ay − az : The computation is similar to that of part a, but we change the direction: dW = −qE · dL = −4(400ax − 300ay + 500az ) · =−

(−2ax + 3ay − az ) (10−3 ) √ 14

(4 × 10−3 ) (−800 − 900 − 500) = 2.35 J √ 14

4.3. If E = 120 aρ V/m, find the incremental amount of work done in moving a 50 µm charge a distance of 2 mm from: a) P (1, 2, 3) toward Q(2, 1, 4): The √ vector along this direction will be Q − P = (1, −1, 1) from which aP Q = [ax − ay + az ]/ 3. We now write   (ax − ay + az −6 dW = −qE · dL = −(50 × 10 ) 120aρ · (2 × 10−3 ) √ 3
 1 = −(50 × 10−6 )(120) (aρ · ax ) − (aρ · ay ) √ (2 × 10−3 ) 3 At P , φ = tan−1 (2/1) = 63.4◦ . Thus (aρ · ax ) = cos(63.4) = 0.447 and (aρ · ay ) = sin(63.4) = 0.894. Substituting these, we obtain dW = 3.1 µJ. b) Q(2, 1, 4) toward P (1, 2, 3): A little thought is in order here: Note that the field has only a radial √ component and does not depend on φ or z. Note also that P and Q are at the same radius ( 5) from the z axis, but have different φ and z coordinates. We could just as well position the two points at the same z location and the problem would not change. If this were so, then moving along a straight line between P and Q would thus involve moving along a chord of a circle whose √ radius is 5. Halfway along this line is a point of symmetry in the field (make a sketch to see this). This means that when starting from either point, the initial force will be the same. Thus the answer is dW = 3.1 µJ as in part a. This is also found by going through the same procedure as in part a, but with the direction (roles of P and Q) reversed. 4.4. Find the amount of energy required to move a 6-C charge from the origin to P (3, 1, −1) in the field E = 2xax − 3y 2 ay + 4az V/m along the straight-line path x = −3z, y = x + 2z: We set up the computation as follows, and find the the result does not depend on the path.   W = −q E · dL = −6 (2xax − 3y 2 ay + 4az ) · (dxax + dyay + dzaz )  1  −1  3 2 2xdx + 6 3y dy − 6 4dz = −24 J = −6 0

0

0

43

P 4.5. Compute the value of A G · dL for G = 2yax with A(1, −1, 2) and P (2, 1, 2) using the path: a) straight-line segments A(1, −1, 2) to B(1, 1, 2) to P (2, 1, 2): In general we would have 

P

A

 G · dL =

P

2y dx

A

The change in x occurs when moving between B and P , during which y = 1. Thus 

P

A

 G · dL =

P

B



2

2y dx =

2(1)dx = 2

1

b) straight-line segments A(1, −1, 2) to C(2, −1, 2) to P (2, 1, 2): In this case the change in x occurs when moving from A to C, during which y = −1. Thus 

P

A

 G · dL =

C

A



2

2y dx =

2(−1)dx = −2

1

 4.6. Let G = 4xax +2zay +2yaz . Given an initial point P (2, 1, 1) and a final point Q(4, 3, 1), find G·dL using the path: a) straight line: y = x − 1, z = 1; b) parabola: 6y = x 2 + 2, z = 1: With G as given, the line integral will be 



4

G · dL =



3

4x dx +

2



1

2z dy +

1

2y dz

1

Clearly, we are going nowhere in z, so the last integral is zero. With z = 1, the first two evaluate as 

3 4   G · dL = 2x 2  + 2y  = 28 2

1

The paths specified in parts a and b did not play a role, meaning that the integral between the specified points is path-independent. 4.7. Repeat Problem 4.6 for G = 3xy 3 ax + 2zay . Now things are different in that the path does matter: a) straight line: y = x − 1, z = 1: We obtain: 

 G · dL =

4



3

3xy dx + 2

2



4

2z dy =

1



3

3x(x − 1) dx + 2

2

2(1) dy = 90

1

b) parabola: 6y = x 2 + 2, z = 1: We obtain: 

 G · dL = 2

4



3

3xy 2 dx +



4

2z dy =

1

2

44

1 x(x 2 + 2)2 dx + 12



3 1

2(1) dy = 82

4.8. A point charge Q1 is located at the origin in free space. Find the work done in carrying a charge Q2 from: (a) B(rB , θB , φB ) to C(rA , θB , φB ) with θ and φ held constant; (b) C(rA , θB , φB ) to D(rA , θA , φB ) with r and φ held constant; (c) D(rA , θA , φB ) to A(rA , θA , φA ) with r and θ held constant: The general expression for the work done in this instance is  W = −Q2

 E · dL = −Q2

Q1 Q1 Q2 ar · (drar + rdθ aθ + r sin θdφaφ ) = − 2 4π0 r 4π0



dr r2

We see that only changes in r will produce non-zero results. Thus for part a we have Q1 Q2 W =− 4π0



rA rB

dr Q1 Q2 = 2 r 4π0



1 1 − rA rB

 J

The answers to parts b and c (involving paths over which r is held constant) are both 0. 4.9. A uniform surface charge density of 20 nC/m2 is present on the spherical surface r = 0.6 cm in free space. a) Find the absolute potential at P (r = 1 cm, θ = 25◦ , φ = 50◦ ): Since the charge density is uniform and is spherically-symmetric, the angular coordinates do not matter. The potential function for r > 0.6 cm will be that of a point charge of Q = 4πa 2 ρs , or V (r) =

0.081 4π(0.6 × 10−2 )2 (20 × 10−9 ) = V with r in meters 4π0 r r

At r = 1 cm, this becomes V (r = 1 cm) = 8.14 V b) Find VAB given points A(r = 2 cm, θ = 30◦ , φ = 60◦ ) and B(r = 3 cm, θ = 45◦ , φ = 90◦ ): Again, the angles do not matter because of the spherical symmetry. We use the part a result to obtain   1 1 VAB = VA − VB = 0.081 − = 1.36 V 0.02 0.03 4.10. Given a surface charge density of 8 nC/m2 on the plane x = 2, a line charge density of 30 nC/m on the line x = 1, y = 2, and a 1-µC point charge at P (−1, −1, 2), find VAB for points A(3, 4, 0) and B(4, 0, 1): We need to find a potential function for the combined charges. That for the point charge we know to be Q Vp (r) = 4π0 r Potential functions for the sheet and line charges can be found by taking indefinite integrals of the electric fields for those distributions. For the line charge, we have  ρl ρl ln(ρ) + C1 dρ + C1 = − Vl (ρ) = − 2π0 ρ 2π0 For the sheet charge, we have  Vs (x) = −

ρs ρs dx + C2 = − x + C2 20 20

45

4.10. (continued) The total potential function will be the sum of the three. Combining the integration constants, we obtain: ρl ρs Q − ln(ρ) − x+C V = 2π0 20 4π0 r The terms in this expression are not referenced to a common origin, since the charges are at different positions. The parameters r, ρ, and x are scalar distances from the charges, and will  √ be treated 2 + (4 − (−1))2 + (−2)2 = = (3 − (−1)) 45, ρA = as such here. For point A we have r A  √ 2 2 (3 − 1) + (4 − 2) = 8, and its distance from the sheet charge is xA = 3 − 2 = 1. The potential at A is then 10−6 8 × 10−9 30 × 10−9 √ VA = ln 8 − (1) + C √ − 2π0 20 4π0 45  √ 2 + (0 − (−1))2 + (1 − 2)2 = 27, At point B, r = (4 − (−1)) B  √ ρB = (4 − 1)2 + (0 − 2)2 = 13, and the distance from the sheet charge is xB = 4 − 2 = 2. The potential at A is then VB =

10−6 8 × 10−9 30 × 10−9 √ ln 13 − (2) + C √ − 2π0 20 4π0 27

Then 10−6 VA − VB = 4π0



  8 × 10−9 8 1 30 × 10−9 1 − ln (1 − 2) = 193 V − √ −√ 2π0 13 20 27 45

4.11. Let a uniform surface charge density of 5 nC/m2 be present at the z = 0 plane, a uniform line charge density of 8 nC/m be located at x = 0, z = 4, and a point charge of 2 µC be present at P (2, 0, 0). If V = 0 at M(0, 0, 5), find V at N(1, 2, 3): We need to find a potential function for the combined charges which is zero at M. That for the point charge we know to be Vp (r) =

Q 4π0 r

Potential functions for the sheet and line charges can be found by taking indefinite integrals of the electric fields for those distributions. For the line charge, we have  Vl (ρ) = −

ρl ρl dρ + C1 = − ln(ρ) + C1 2π0 ρ 2π0

For the sheet charge, we have  Vs (z) = −

ρs ρs dz + C2 = − z + C2 20 20

The total potential function will be the sum of the three. Combining the integration constants, we obtain: ρl ρs Q − ln(ρ) − z+C V = 4π0 r 2π0 20

46

4.11. (continued) The terms in this expression are not referenced to a common origin, since the charges are at different positions. The parameters r, ρ, and z are scalar distances from the charges, and will be treated √ as such here. √ To evaluate the constant, C, we first look at point M, where VT = 0. At M, r = 22 + 52 = 29, ρ = 1, and z = 5. We thus have 0=

5 × 10−9 2 × 10−6 8 × 10−9 ln(1) − 5 + C ⇒ C = −1.93 × 103 V √ − 2π 2 4π0 29 0 0

At point N, r = VN =

√ √ √ 1 + 4 + 9 = 14, ρ = 2, and z = 3. The potential at N is thus

√ 2 × 10−6 5 × 10−9 8 × 10−9 ln( 2) − (3) − 1.93 × 103 = 1.98 × 103 V = 1.98 kV √ − 2π0 20 4π0 14

4.12. Three point charges, 0.4 µC each, are located at (0, 0, −1), (0, 0, 0), and (0, 0, 1), in free space. a) Find an expression for the absolute potential as a function of z along the line x = 0, y = 1: From  a point located at position  line, the distances to the three charges are √ z along the given R1 = (z − 1)2 + 1, R2 = z2 + 1, and R3 = (z + 1)2 + 1. The total potential will be q V (z) = 4π0



1 1 1 + + R1 R2 R3



Using q = 4 × 10−7 C, this becomes

1

1

1

+√ + V (z) = (3.6 × 103 )  z2 + 1 (z − 1)2 + 1 (z + 1)2 + 1

V

b) Sketch V (z). The sketch will show that V maximizes to a value of 8.68 × 103 at z = 0, and then monotonically decreases with increasing |z| symmetrically on either side of z = 0. 4.13. Three identical point charges of 4 pC each are located at the corners of an equilateral triangle 0.5 mm on a side in free space. How much work must be done to move one charge to a point equidistant from the other two and on the line joining them? This will be the magnitude of the charge times the potential difference between the finishing and starting positions, or W =

(4 × 10−12 )2 2π0



 1 1 − × 104 = 5.76 × 10−10 J = 576 pJ 2.5 5

4.14. two 6-nC point charges are located at (1, 0, 0) and (−1, 0, 0) in free space. a) Find V at P (0, 0, z): Since the charges are positioned symmetrically about the z axis, the potential at z will be double that from one charge. This becomes: V (z) = (2)

q q = √ √ 2 4π0 z + 1 2π0 z2 + 1

b) Find Vmax : It is clear from the part a result that V will maximize at z = 0, or vmax = q/(2π 0 ) = 108 V. 47

4.14. (continued) c) Calculate |dV /dz| on the z axis: Differentiating the part a result, we find  dV  qz   V/m =  dz π0 (z2 + 1)3/2 d) Find |dV /dz|max : To find this we need to differentiate the part c result and find its zero: d  dV  q(1 − 2z2 ) 1 = 0 ⇒ z = ± √  = dz dz π0 (z2 + 1)5/2 2 √ Substituting z = 1/ 2 into the part c result, we find  dV  q   =√ = 83.1 V/m   dz max 2π0 (3/2)3/2 4.15. Two uniform line charges, 8 nC/m each, are located at x = 1, z = 2, and at x = −1, y = 2 in free space. If the potential at the origin is 100 V, find V at P (4, 1, 3): The net potential function for the two charges would in general be: V =− At the origin, R1 = R2 =

ρl ρl ln(R1 ) − ln(R2 ) + C 2π0 2π0

√ 5, and V = 100 V. Thus, with ρl = 8 × 10−9 ,

100 = −2

(8 × 10−9 ) √ ln( 5) + C ⇒ C = 331.6 V 2π0

At P (4, 1, 3), R1 = |(4, 1, 3) − (1, 1, 2)| =

√ √ 10 and R2 = |(4, 1, 3) − (−1, 2, 3)| = 26. Therefore

√  (8 × 10−9 )  √ VP = − ln( 10) + ln( 26) + 331.6 = −68.4 V 2π0

48

4.16. Uniform surface charge densities of 6, 4, and 2 nC/m2 are present at r = 2, 4, and 6 cm, respectively, in free space. a) Assume V = 0 at infinity, and find V (r). We keep in mind the definition of absolute potential as the work done in moving a unit positive charge from infinity to location r. At radii outside all three spheres, the potential will be the same as that of a point charge at the origin, whose charge is the sum of the three sphere charges: [4π(.02)2 (6) + 4π(.04)2 (4) + 4π(.06)2 (2)] × 10−9 q 1 + q2 + q3 = 4π0 r 4π0 r −13 (96 + 256 + 288)π × 10 1.81 = = V where r is in meters 4π(8.85 × 10−12 )r r

V (r) (r > 6 cm) =

As the unit charge is moved inside the outer sphere to positions 4 < r < 6 cm, the outer sphere contribution to the energy is fixed at its value at r = 6. Therefore, V (r) (4 < r < 6 cm) =

q1 + q2 q3 0.994 + = + 13.6 V 4π0 r 4π0 (.06) r

In moving inside the sphere at r = 4 cm, the contribution from that sphere becomes fixed at its potential function at r = 4: V (r) (2 < r < 4 cm) =

q1 q2 q3 0.271 + + = + 31.7 V 4π0 r 4π0 (.04) 4π0 (.06) r

Finally, using the same reasoning, the potential inside the inner sphere becomes V (r) (r < 2 cm) =

0.271 + 31.7 = 45.3 V .02

b) Calculate V at r = 1, 3, 5, and 7 cm: Using the results of part a, we substitute these distances (in meters) into the appropriate formulas to obtain: V (1) = 45.3 V, V (3) = 40.7 V, V (5) = 33.5 V, and V (7) = 25.9 V. c) Sketch V versus r for 0 < r < 10 cm.

49

4.17. Uniform surface charge densities of 6 and 2 nC/m2 are present at ρ = 2 and 6 cm respectively, in free space. Assume V = 0 at ρ = 4 cm, and calculate V at: a) ρ = 5 cm: Since V = 0 at 4 cm, the potential at 5 cm will be the potential difference between points 5 and 4:    5  5 (.02)(6 × 10−9 ) aρsa 5 dρ = − = −3.026 V E · dL = − ln V5 = − 0 4 4 4 0 ρ b) ρ = 7 cm: Here we integrate piecewise from ρ = 4 to ρ = 7: 

6

V7 = − 4

aρsa dρ − 0 ρ



7 6

(aρsa + bρsb ) dρ 0 ρ

With the given values, this becomes         6 7 (.02)(6 × 10−9 ) (.02)(6 × 10−9 ) + (.06)(2 × 10−9 ) V7 = − ln ln − 0 4 0 6 = −9.678 V 4.18. A nonuniform linear charge density, ρL = 8/(z2 + 1) nC/m lies along the z axis. Find the potential at P (ρ = 1, 0, 0) in free space if V = 0 at infinity: This last condition enables us to write the potential at P as a superposition of point charge potentials. The result is the integral:  ∞ ρL dz VP = −∞ 4π0 R √ where R = z2 + 1 is the distance from a point z on the z axis to P . Substituting the given charge distribution and R into the integral gives us  VP =

∞

−∞

∞ 2 × 10−9 8 × 10−9 dz z  = = 144 V √  2 4π0 (z2 + 1)3/2 π0 z + 1 −∞

4.19. The annular surface, 1 cm < ρ < 3 cm, z = 0, carries the nonuniform surface charge density ρs = 5ρ nC/m2 . Find V at P (0, 0, 2 cm) if V = 0 at infinity: We use the superposition integral form:   ρs da VP = 4π0 |r − r | where r = zaz and r = ρaρ . We integrate over the surface of the annular region, with da = ρ dρ dφ. Substituting the given values, we find  VP =

0

2π



.03

.01

(5 × 10−9 )ρ 2 dρ dφ  4π0 ρ 2 + z2

Substituting z = .02, and using tables, the integral evaluates as 

(5 × 10−9 ) VP = 20

  .03  ρ (.02)2 2 2 2 2 ln(ρ + ρ + (.02) ) ρ + (.02) − = .081 V 2 2 .01 50

4.20. Fig. 4.11 shows three separate charge distributions in the z = 0 plane in free space. a) find the total charge for each distribution: Line charge along the y axis:  5 π × 10−9 dy = 2π × 10−9 C = 6.28 nC Q1 = 3

Line charge in an arc at radius ρ = 3:  Q2 =

70◦

10◦

(10−9 ) 3 dφ = 4.5 × 10−9 (70 − 10)

2π = 4.71 × 10−9 C = 4.71 nC 360

Sheet charge:  Q3 =



70◦

10◦

3.5

(10−9 ) ρ dρ dφ = 5.07 × 10−9 C = 5.07 nC

1.6

b) Find the potential at P (0, 0, 6) caused by each of the three charge distributions acting alone: Line charge along y axis:  5  5  5 ρL dL π × 10−9 dy 103  = ln(y + y 2 + 62 ) = 7.83 V = VP 1 =  2 2 3 4π R 4 × 8.854 0 3 3 4π0 y + 6 Line charge in an arc a radius ρ = 3:  VP 2 =

70◦

10◦

(1.5 × 10−9 ) 3 dφ Q2 = √ √ = 6.31 V 4π0 45 4π0 32 + 62

Sheet charge: 

70◦



(10−9 ) ρ dρ dφ 60 × 10−9 =  4π(8.854 × 10−12 10◦ 1.6 4π0 ρ 2 + 62   3.5 = 9.42 ρ 2 + 36 = 6.93 V

VP 3 =

3.5



2π 360



3.5

1.6

ρ dρ  ρ 2 + 36

1.6

c) Find VP : This will be the sum of the three results of part b, or VP = VP 1 + VP 2 + VP 3 = 7.83 + 6.31 + 6.93 = 21.1 V 4.21. Let V = 2xy 2 z3 + 3 ln(x 2 + 2y 2 + 3z2 ) V in free space. Evaluate each of the following quantities at P (3, 2, −1): a) V : Substitute P directly to obtain: V = −15.0 V b) |V |. This will be just 15.0 V. c) E: We have       6x 12y   2 3 3 ax + 4xyz + 2 ay E = −∇V  = − 2y z + 2 P P x + 2y 2 + 3z2 x + 2y 2 + 3z2    18z + 6xy 2 z2 + 2 = 7.1ax + 22.8ay − 71.1az V/m az x + 2y 2 + 3z2 P 51

4.21d. |E|P : taking the magnitude of the part c result, we find |E|P = 75.0 V/m. e) aN : By definition, this will be  E  = −0.095 ax − 0.304 ay + 0.948 az aN  = − P |E|     f) D: This is D = 0 E = 62.8 ax + 202 ay − 629 az pC/m2 . P

P

4.22. It is known that the potential is given as V = 80r 0.6 V. Assuming free space conditions, find: a) E: We use dV E = −∇V = − ar = −(0.6)80r −0.4 ar = −48r −0.4 ar V/m dr b) the volume charge density at r = 0.5 m: Begin by finding D = 0 E = −48r −0.4 0 ar C/m2 We next find ρv = ∇ · D =

 1 d  2  1 d  76.80 1.6 r −48 D r = = − 1.4 C/m3 r 0 r 2 dr r 2 dr r

Then at r = 0.5 m, ρv (0.5) =

−76.8(8.854 × 10−12 ) = −1.79 × 10−9 C/m3 = −1.79 nC/m3 (0.5)1.4

c) the total charge lying within the surface r = 0.6: The easiest way is to use Gauss’law, and integrate the flux density over the spherical surface r = 0.6. Since the field is constant at constant radius, we obtain the product: Q = 4π(0.6)2 (−480 (0.6)−0.4 ) = −2.36 × 10−9 C = −2.36 nC 4.23. It is known that the potential is given as V = 80ρ .6 V. Assuming free space conditions, find: a) E: We find this through E = −∇V = −

dV aρ = −48ρ −.4 V/m dρ

b) the volume charge density at ρ = .5 m: Using D = 0 E, we find the charge density through       1 d    ρv  = [∇ · D].5 = ρDρ  = −28.80 ρ −1.4  = −673 pC/m3 .5 .5 .5 ρ dρ

52

4.23c. the total charge lying within the closed surface ρ = .6, 0 < z < 1: The easiest way to do this calculation is to evaluate D ρ at ρ = .6 (noting that it is constant), and then multiply by the cylinder area: Using part  a, we have Dρ  = −480 (.6)−.4 = −521 pC/m2 . Thus Q = −2π(.6)(1)521×10−12 C = −1.96 nC. .6

4.24. Given the potential field V = 80r 2 cos θ and a point P (2.5, θ = 30◦ , φ = 60◦ ) in free space, find at P : a) V : Substitute the coordinates into the function and find VP = 80(2.5)2 cos(30) = 433 V. b) E:

∂V 1 ∂V ar − aθ = −160r cos θar + 80r sin θaθ V/m ∂r r ∂θ Evaluating this at P yields Ep = −346ar + 100aθ V/m. E = −∇V = −

c) D: In free space, DP = 0 EP = (−346ar + 100aθ )0 = −3.07 ar + 0.885 aθ nC/m2 . d) ρv :

 ρv = ∇ · D = 0 ∇ · E = 0

 ∂ 1 ∂  2  1 r Er + 2 (Eθ sin θ ) r 2 ∂r r sin θ ∂θ

Substituting the components of E, we find   160 cos θ 2 1 ρv = − 80r(2 sin θ cos θ) 0 = −3200 cos θ = −2.45 nC/m3 3r + r2 r sin θ with θ = 30◦ . e) dV /dN: This will be just |E| evaluated at P , which is  dV   = | − 346ar + 100aθ | = (346)2 + (100)2 = 360 V/m dN P f) aN : This will be

EP −346ar + 100aθ aN = − =−  = 0.961 ar − 0.278 aθ |EP | (346)2 + (100)2 4.25. Within the cylinder ρ = 2, 0 < z < 1, the potential is given by V = 100 + 50ρ + 150ρ sin φ V. a) Find V , E, D, and ρv at P (1, 60◦ , 0.5) in free space: First, substituting the given point, we find VP = 279.9 V. Then, E = −∇V = −

∂V 1 ∂V aρ − aφ = − [50 + 150 sin φ] aρ − [150 cos φ] aφ ∂ρ ρ ∂φ

Evaluate the above at P to find EP = −179.9aρ − 75.0aφ V/m Now D = 0 E, so DP = −1.59aρ − .664aφ nC/m2 . Then      1 ∂Dφ 1 1 d  50 1 ρv = ∇ · D = = − (50 + 150 sin φ) + 150 sin φ 0 = − 0 C ρDρ + ρ dρ ρ ∂φ ρ ρ ρ At P , this is ρvP = −443 pC/m3 . 53

4.25b. How much charge lies within the cylinder? We will integrate ρv over the volume to obtain:  Q= 0

1  2π



0

2 0

−

500 ρ dρ dφ dz = −2π(50)0 (2) = −5.56 nC ρ

4.26. A dipole having Qd/(4π0 ) = 100 V · m2 is located at the origin in free space and aligned so that its moment is in the az direction. a) Sketch |V (r = 1, θ, φ = 0)| versus θ on polar graph paper (homemade if you wish). b) Sketch |E(r = 1, θ, φ = 0)| versus θ on polar graph paper: V =

Qd cos θ 100 cos θ = ⇒ |V (r = 1, θ, φ = 0)| = |100 cos θ| 4π0 r 2 r2

Qd 100 (2 cos θ ar + sin θ aθ ) = 3 (2 cos θ ar + sin θ aθ ) 3 4π0 r r 1/2  1/2  = 100 1 + 3 cos2 θ |E(r = 1, θ, φ = 0)| = 100 4 cos2 θ + sin2 θ E=

These results are plotted below:

54

4.27. Two point charges, 1 nC at (0, 0, 0.1) and −1 nC at (0, 0, −0.1), are in free space. a) Calculate V at P (0.3, 0, 0.4): Use VP =

q q − + 4π0 |R | 4π0 |R− |

where R+ = (.3, 0, .3) and R− = (.3, 0, .5), so that |R+ | = 0.424 and |R− | = 0.583. Thus VP =

10−9 4π0



 1 1 − = 5.78 V .424 .583

b) Calculate |E| at P : Use EP =

 q(.3ax + .3az ) q(.3ax + .5az ) 10−9
− = + 1.41a 2.42a V/m x z 4π0 (.424)3 4π0 (.583)3 4π0

Taking the magnitude of the above, we find |EP | = 25.2 V/m. c) Now treat the two√charges as a dipole at the origin and find V at P : In spherical coordinates, P is located at r = .32 + .42 = .5 and θ = sin−1 (.3/.5) = 36.9◦ . Assuming a dipole in far-field, we have qd cos θ 10−9 (.2) cos(36.9◦ ) VP = = = 5.76 V 4π0 r 2 4π0 (.5)2 4.28. A dipole located at the origin in free space has a moment p2 × 10−9 az C · m. At what points on the line y = z, x = 0 is: a) |Eθ | = 1 mV/m? We note that the line y = z lies at θ = 45◦ . Begin with E=

2 × 10−9 10−9 (2 cos θ a + sin θ a ) = (2ar + aθ ) at θ = 45◦ √ r θ 3 4π0 r 3 2 2π0 r

from which Eθ =

10−9 = 10−3 V/m (required) ⇒ r 3 = 1.27 × 10−4 or r = 23.3 m 2π0 r 3

√ The y and z values are thus y = z = ±23.3/ 2 = ±16.5 m b) |Er | = 1 mV/m? From the above field expression, the radial component magnitude is twice that of the theta component. Using the same development, we then find Er = 2

10−9 = 10−3 V/m (required) ⇒ r 3 = 2(1.27 × 10−4 ) or r = 29.4 m 2π0 r 3

√ The y and z values are thus y = z = ±29.4/ 2 = ±20.8 m

55

4.29. A dipole having a moment p = 3ax − 5ay + 10az nC · m is located at Q(1, 2, −4) in free space. Find V at P (2, 3, 4): We use the general expression for the potential in the far field: V =

p · (r − r ) 4π0 |r − r |3

where r − r = P − Q = (1, 1, 8). So VP =

(3ax − 5ay + 10az ) · (ax + ay + 8az ) × 10−9 = 1.31 V 4π0 [12 + 12 + 82 ]1.5

4.30. A dipole, having a moment p = 2az nC · m is located at the origin in free space. Give the magnitude of E and its direction aE in cartesian components at r = 100 m, φ = 90◦ , and θ =: a) 0◦ ; b) 30◦ ; c) 90◦ . Begin with p E= [2 cos θ ar + sin θ aθ ] 4π0 r 3 from which |E| =

1/2 1/2 p  p  2 2 2 4 cos 1 + 3 cos θ + sin θ = θ 4π0 r 3 4π0 r 3

Now E x = E · ax =

p p [2 cos θ ar · ax + sin θ aθ · ax ] = [3 cos θ sin θ cos φ] 3 4π0 r 4π0 r 3

Ey = E · ay =

 p p
2 cos θ ar · ay + sin θ aθ · ay = [3 cos θ sin θ sin φ] 3 4π0 r 4π0 r 3

Ez = E · az =

  p  p
2 2 2 cos θ a = · a + sin θ a · a θ − sin θ 2 cos r z θ z 4π0 r 3 4π0 r 3

then

and

Since φ is given as 90◦ , Ex = 0, and the field magnitude becomes |E(φ = 90◦ )| =



Ey2 + Ez2 =

1/2 p  2 2 2 2 2 θ sin θ + (2 cos θ − sin θ) 9 cos 4π0 r 3

Then the unit vector becomes (again at φ = 90◦ ): 3 cos θ sin θ ay + (2 cos2 θ − sin2 θ) az 1/2 9 cos2 θ sin2 θ + (2 cos2 θ − sin2 θ)2

aE =


Now with r = 100 m and p = 2 × 10−9 , p 2 × 10−9 = = 1.80 × 10−5 4π0 r 3 4π(8.854 × 10−12 )106 Using the above formulas, we find at θ = 0◦ , |E| = (1.80 × 10−5 )(2) = 36.0 µV/m and aE = az . At θ = 30◦ , we find |E| = (1.80 × 10−5 )[1.69 + 1.56]1/2 = 32.5 µV/m and aE = (1.30ay + 1.25az )/1.80 = 0.72 ax + 0.69 az . At θ = 90◦ , |E| = (1.80×10−5 )(1) = 18.0 µV/m and aE = −az . 56

4.31. A potential field in free space is expressed as V = 20/(xyz) V. a) Find the total energy stored within the cube 1 < x, y, z < 2. We integrate the energy density over the cube volume, where wE = (1/2)0 E · E, and where 

 1 1 1 E = −∇V = 20 2 ax + 2 ay + az V/m x yz xy z xyz2 The energy is now 2 2 2

 WE = 2000

1

1

1

1 1 1 + 2 4 2+ 2 2 4 4 2 2 x y z x y z x y z

 dx dy dz

The integral evaluates as follows:   2 1 1 1 1 − − 4 2 − 2 4 dy dz WE = 2000 3 x 3 y 2 z2 xy z xy z 1 1 1       2  2   1 1 1 7 1 1 dy dz + + = 2000 24 y 2 z2 2 y 4 z2 2 y 2 z4 1 1       2   1 1 1 2 7 1 1 − − − dz = 2000 24 yz2 6 y 3 z2 2 yz4 1 1       2   1 1 7 7 1 1 dz + + = 2000 2 2 48 z 48 z 4 z4 1   7 = 387 pJ = 2000 (3) 96 

2 2

b) What value would be obtained by assuming a uniform energy density equal to the value at the center of the cube? At C(1.5, 1.5, 1.5) the energy density is  1 wE = 2000 (3) = 2.07 × 10−10 J/m3 (1.5)4 (1.5)2 (1.5)2 

This, multiplied by a cube volume of 1, produces an energy value of 207 pJ. 4.32. In the region of free space where 2 < r < 3, 0.4π < θ < 0.6π, 0 < φ < π/2, let E = k/r 2 ar . a) Find a positive value for k so that the total energy stored is exactly 1 J: The energy is found through  π/2  0.6π  3 1 1 k2 2 2 0 E dv = 0 2 r sin θ dr dθ dφ WE = v 2 0 0.4π 2 2 r   .6π  1  π 1 3 0.616π  2 = (− cos θ) 0 k − 0 k 2 = 1 J  = .4π 2 2 r 2 24 

Solve for k to find k = 1.18 × 106 V · m.

57

4.32b. Show that the surface θ = 0.6π is an equipotential surface: This will be the surface of a cone, centered at the origin, along which E, in the ar direction, will exist. Therefore, the given surface cannot be an equipotential (the problem was ill-conceived). Only a surface of constant r could be an equipotential in this field. c) Find VAB , given points A(2, θ = π/2, φ = π/3) and B(3, π/2, π/4): Use  VAB = −

A B



3

E · dL = − 2

k ar · ar dr = k r2



1 1 − 2 3

 =

k 6

Using the result of part a, we find VAB = (1.18 × 106 )/6 = 197 kV. 4.33. A copper sphere of radius 4 cm carries a uniformly-distributed total charge of 5 µC in free space. a) Use Gauss’ law to find D external to the sphere: with a spherical Gaussian surface at radius r, D will be the total charge divided by the area of this sphere, and will be ar -directed. Thus Q 5 × 10−6 a = ar C/m2 r 4πr 2 4πr 2

D=

b) Calculate the total energy stored in the electrostatic field: Use  2π  π  ∞ 1 1 (5 × 10−6 )2 2 D · E dv = r sin θ dr dθ dφ 2 4 vol 2 0 0 .04 2 16π 0 r    1 (5 × 10−6 )2 ∞ dr 25 × 10−12 1 = (4π) = 2.81 J = 2 2 16π 2 0 8π0 .04 .04 r 

WE =

c) Use WE = Q2 /(2C) to calculate the capacitance of the isolated sphere: We have Q2 (5 × 10−6 )2 = 4.45 × 10−12 F = 4.45 pF = 2WE 2(2.81)

C=

4.34. Given the potential field in free space, V = 80φ V (note that ap hi should not be present), find: a) the energy stored in the region 2 < ρ < 4 cm, 0 < φ < 0.2π, 0 < z < 1 m: First we find E = −∇V = −

1 dV 80 aφ = − aφ V/m ρ dφ ρ

Then  WE =

v

1  0.2π

 wE dv =

0

0



.04

.02

  1 (80)2 .04 0 2 ρ dρ dφ dz = 640π0 ln = 12.3 nJ 2 ρ .02

b) the potential difference, VAB , for A(3 cm, φ = 0, z = 0) and B(3cm, 0.2π, 1m): Use  VAB = −

A B

 E · dL = −

0 .2π

−

80 aφ · aφ ρ dφ = −80(0.2π) = −16π V ρ

58

4.34c. the maximum value of the energy density in the specified region: The energy density is wE =

1 1 6400 0 E 2 = 0 2 2 2 ρ

This will maximize at the lowest value of ρ in the specified range, which is ρ = 2 cm. So wE,max =

1 6400 0 = 7.1 × 10−5 J/m3 = 71 µJ/m3 2 .022

4.35. Four 0.8 nC point charges are located in free space at the corners of a square 4 cm on a side. a) Find the total potential energy stored: This will be given by 1 q n Vn 2 4

WE =

n=1

where Vn in this case is the potential at the location of any one of the point charges that arises from the other three. This will be (for charge 1) V1 = V21 + V31 + V41

q = 4π0



1 1 1 + + √ .04 .04 .04 2



Taking the summation produces a factor of 4, since the situation is the same at all four points. Consequently,   1 1 (.8 × 10−9 )2 2 + √ = 7.79 × 10−7 J = 0.779 µJ WE = (4)q1 V1 = 2 2π0 (.04) 2 b) A fifth 0.8 µC charge is installed at the center of the square. Again find the total stored energy: This will be the energy found in part a plus the amount of work done in moving the fifth charge into position from infinity. The latter is just the potential at the square center arising from the original four charges, times the new charge value, or 2WE =

4(.8 × 10−9 )2 = .813 µJ √ 4π0 (.04 2/2)

The total energy is now WE net = WE (part a) + 2WE = .779 + .813 = 1.59 µJ

59